Let's take Question A:
Both fathers have siblings with red ears, and red ears are an autosomal recessive trait. The grandparents did not have red ears, we know then that they were carriers of the recessive allele. Each grandparent was Nr (for Normal and red alleles respectively).
The fathers have normal ears, so they could be NN (probability 0.33) or Nr (probability 0.67).
The answer then is that each father independently has a 2/3 chance (0.67 probability) of carrying the red ears gene.
The male's mother, being RR, makes the male a carrier only if the male's father is himself a carrier. To put it mathematically, he has a 1/2 chance of being a carrier with 2/3 probability that he was a carrier to begin with. (2/3)*(1/2) = 1/3.
Therefore the male can be carrier with 1/3 chance.
Now Question B:
Since the male and female have symmetrical pedigrees, and we just solved the chance of the male being a carrier with 1/3 chance, then the chance of them having a child with red ears is (1/3)*(1/3) = 1/9.
The male's and female's fathers may each be heterozygous or homozygous normal, given the information, with both mothers being homozygous normal (NN).
If Father A is NN (crossed with Mother A, also NN), there is a 0 probability of passing on the red ears allele.
If Father A is Nr, then Person A has a 0.5 probability of carrying the red ear allele.
Person A can now be NN (with 0.25 probability) or Nr (with 0.5 probability).
Likewise with Person B.
Now lets consider all 4 combinations of genotypes that Person A and B can have when they have a child. I will write a table with one Person on each side of the table, with the probability of their genotype written in brackets beside. Each intersection will represent the frequency of carrying the red ears allele.
Person A NN Nr
(1/3) NN 0/4 2/4
(2/3) Nr 2/4 2/4
Carrying the red ear gene: (A more interesting example)
There are a total of 6 outcomes in which the child carries the red ears allele, but these must be weighted by the probability that each parent has the associated genotype.
Probability of carrying 'r' = [ (1/3)(1/3)(0/4) + (1/3)(2/3)(2/4) + (2/3)(1/3)(2/4) + (2/3)(2/3)(2/4) ] / [ (1/3)(1/3) + (1/3)(2/3) + (2/3)(1/3) + (2/3)(2/3) ]
Having a child with red ears:
Probability of having red ears = [ (1/3)(1/3)(0/4) + (1/3)(2/3)(0/4) + (2/3)(1/3)(0/4) + (2/3)(2/3)(1/4) ] / [ (1/3)(1/3) + (1/3)(2/3) + (2/3)(1/3) + (2/3)(2/3) ]
= 1/9 or about 0.11
It's left as an exercise to the student to derive the remaining probabilites.